Proving $\nabla F(x,y,z)$ is normal to the surface $F(x,y,z)=0$ What would be a simple way to prove that $\nabla F(x,y,z)$ is normal to the surface $F(x,y,z)=0$? I was wondering if anyone had a simple way to do this.
Thanks in advance 
 A: Let $u,v$ parametrise the surface $S$ where $F(x,y,z) = 0$ on $S$. Then we can write $F(S(u,v)) = 0$. What happens when we take derivatives with respect to $u$ and $v$?
(Hint: apply chain rule)
Edit: Taking partial derivatives of $F$ over the surface with respect to, for example, u, gives $\frac{\partial F(S(u,v))}{\partial u} = \frac{\partial F}{\partial x_i} \frac{\partial x_i}{\partial u}$. $\frac{\partial F}{\partial x_i} = \nabla F$,  and since this derivative is along the curve, $\frac{\partial x_i}{\partial u} = \frac{\partial S}{\partial u} $. So we're left with $\frac{\partial F(S(u,v))}{\partial u} =\nabla F \bullet  \frac{\partial S}{\partial u} $. 
But since this is along the surface, the left hand side is zero (since changing u keeps us on the surface, so $F$ doesn't change.) so $\nabla F$ is normal to $\frac{\partial S}{\partial u}$. Similarly for $\frac{\partial S}{\partial v}$. But the partial derivatives are the two tangent vectors that span the plane normal to the surface, so $\nabla F$ mus be normal to the plane!
Note that this result is true for $F(x,y,z) = c$ on a surface for any constant $c$, not just $0$.
